Tag Info

New answers tagged

3

First of all, we can assume that $b=1$ without loss of generality. For, if we define $\overline{a}=\frac{a}{b}$, then minimizing $|\overline{a}z_x-z_y|$ is the same as minimizing $|az_x-bz_y|$. Now, we're really looking for $\frac{z_y}{z_x}$ to be a rational number close to $a$. The convergents of the continued fraction expansion of $a$ are precisely the ...


0

It's $1/(e^{1/2}-1)$. You should be able to derive this by doing a term-by-term transformation on an appropriate infinite series.


1

In general, for any positive real number $x$ you get the continued fraction as follows. Let $\lfloor x \rfloor = a_0$. Then $x = a_0 + \dfrac{1}{r_1}$ where $r_1 = \dfrac{1}{x - a_0}$. Continue with $\lfloor r_1 \rfloor = a_1$, and $r_2 = \dfrac{1}{r - a_1}$, etc. Then $$x = a_0 + \dfrac{1}{a_1 + \dfrac{1}{a_2 + \ldots}}$$ In this case $$\eqalign{x ...


0

If we set $x = \dfrac{1}{-1 +\underbrace{\dfrac{1}{-1+\dfrac{1}{-1+\dfrac{1}{-1+\cdots}}}}_{x}}$, then we have $x = \frac{1}{-1 + x}$, which gives also $x^2 - x - 1=0$ That's a similar presentation of the negative answer.


10

I think it's obviously a negative number of magnitude less than 1. After all, $1 / x$ is clearly a negative number of magnitude greater than $1$! Okay not really. But the reason you think it's obviously positive is that there is a hidden condition in your question: you intend $x$ to be the limit of the sequence $$1, 1 + \frac{1}{1}, 1 + \frac{1}{1 + ...


1

Let $f(z)=1+\frac1z$. The function $f$ has two fixed points, which are $\frac{1\pm\sqrt{5}}{2}$. At first, when you write $$x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}}}$$ you're defining $x$ to be the particular fixed point of $f$ that attracts the orbit of $1$, if it exists. It so happens that $x$ does exist, and it equals ...



Top 50 recent answers are included